googologywikiaorg-20200223-history
User talk:Boboris02/SON analysis
The limit of \(\alpha -> \sigma(\sigma(\alpha),0)\) is \(\psi(\Omega_{\omega})\). Googleaarex (talk) 16:15, February 22, 2017 (UTC) Aarex,you are quite wrong.I will continue the analysis whenever I can,but your guess is quite far from reality.The relation \(\forall \alpha,\beta (f_{\alpha}(n) \approx g_{\beta}(n)):\sigma(\alpha) = \beta\) stops after \(\varepsilon_0\).Please try not to make bold statements like that.Only because a relation holds for SOME group of ordinals,doesn't mean it will hold for ANY ordinal.Boboris02 (talk) 18:15, February 26, 2017 (UTC) Would you mind giving some examples of \(L(\alpha)\) and their suprema as well? I still struggle to understand what they are. LittlePeng9 (talk) 17:10, February 22, 2017 (UTC) It is quite dificult.I understand your confusion.I didn't really takle this too much in the blog post. First thing you need to know is:\(L(0) = 1\) Now you can look up my definition in the blog.I said that \(L(\alpha)\) is the set of all ordinals expressable using \(\sigma\) no more than \(\alpha\) times and without using ordinals that are not in the set \(L(\delta)\) for all \(\delta +1 \geq \alpha\).Let's try to decipher this.What is \(L(1)\)?Earlier I said \(L(0)\) is one and given that in this case \(\alpha = 1\) then we can only use \(\sigma\) one time and we cant use ordinals that are not in the set \(L(\delta)\) if \(\delta +1 \geq \alpha = 1\).That means we can only use 1.We can then use \(\sigma(0,1)\),which is equal to 2.Any previously determined ordinal in the set is less than another ordinal in the same set \(L(\alpha)\) if it can be in the definition of that ordinal within \(\alpha\) or less \(\sigma\)'s.So then,\(\sigma(0,2)\),which is 3 is in \(L(1)\).Then we use 3 in \(\sigma(0,3)\) to generate 4 and so on.As you can see we get to the first fixed point of \(\alpha \mapsto 1 + \alpha\),which is \(\omega\).The supremum of \(L(1)\) is \(\omega\) and generally the supremum to \(L(\alpha +1)\) is \(\sigma(L(\alpha))\).Boboris02 (talk) 18:15, February 26, 2017 (UTC) :I disagree with some of the things you say. Firstly, I don't see why \(L(0)=\{1\}\) (I assume this is what you mean, since \(L(0)\) is a set), but I can agree to this one for now (so that I can get to other things). :To find elements of the set \(L(1)\) you are allowed to use only elements of the \(L(\delta)\) for \(\delta+1\geq 1\), so elements of \(L(\delta)\) for any \(\delta\). But you seem to be also using \(0\), and in no way it is justified that \(0\) is an element of any \(L(\delta)\). On the other hand, you seem to be ignoring that using just one \(\sigma\) and even just the elements of \(L(0)\), you can also form the number \(\sigma(1,1)=\omega 2\). Since we already know \(k<\omega\) are in \(L(1)\), we can also form \(\sigma(k,l)\), which gives us some very large ordinals. I won't even mention that we can plug these large ordinals (which we now know are in \(L(1)\)) into \(\sigma\). This makes the supremum of \(L(1)\) be a very large ordinal, not \(\omega\). :Also, though this is a minor issue, I suppose you meant to say that supremum of \(L(\alpha+1)\) is \(\sigma(\sup(L(\alpha)))\) (since again, \(L(\alpha)\) is a set of ordinals, not an ordinal itself). LittlePeng9 (talk) 19:03, February 26, 2017 (UTC) :I agree.I think if I change the definition to \(L(0)=\{0,1\}\) the zero problem should desapear.However,you are wrong about the \(\sigma(k,l)\) statement.You cannot use \(\sigma(1,1)\) unless you can express \(\sigma(0,\sigma(1))\) and \(\sigma(0,\sigma(0,(\sigma(1))))\) ect. and you can't have \(\sigma(0,\sigma(1,0))\),because \(\sigma(1)\) is already a supremum to the set \(L(1)\).Boboris02 (talk) 19:14, February 26, 2017 (UTC) ::Nothing in the definition of the \(L(\alpha)\) indicates that, in order to consider something (in our case, \(\sigma(1,1)\)) to be an element of it, we have to consider expressibility of some other ordinals in the desired form. ::I am going to split my argument into small steps, and I want you to indicate with which steps you don't agree. ::1. \(1\in L(0)\). ::2. Hence, for \(\alpha=1,\gamma=1\) and \(\delta=0\), we have \(\delta+1\leq\alpha\) and \(\gamma\in L(\delta)\) ::3. Therefore, for \(\alpha=1,\gamma=1\) it is not true that \(\gamma\not\in L(\delta)\forall\delta(\delta+1\leq\alpha)\). ::4. Expression \(\sigma(1,1)\) uses \(\sigma\) no more than \(\alpha=1\) times and it uses no ordinal \(\gamma\) if \(\gamma\not\in L(\delta)\forall\delta(\delta+1\leq\alpha)\) (since it uses only \(\gamma=1\), which by 3. doesn't satisfy this condition). ::5. Thus \(\beta=\sigma(1,1)\) is definable using \(\sigma\) no more than \(\alpha=1\) times and it uses no ordinal \(\gamma\) if \(\gamma\not\in L(\delta)\forall\delta(\delta+1\leq\alpha)\). ::6. So, by definition of \(L(\alpha)\), \(\beta\in L(\alpha)\) for \(\alpha=1\) and \(\beta=\sigma(1,1)\). ::The first step is what you have stated above, and the sixth is just the application of the definition. The other four are, in my opinion, all logical deductions, but since you disagree with the last conclusion, one of them must be wrong. Which one? LittlePeng9 (talk) 19:38, February 26, 2017 (UTC) ::You are wrong at 3 because that's not what I meant.It's not whether the statement is true or not.The context was about how \(L(\alpha)\) works.It said "......with using \(\sigma\) no more than \(\alpha\) times and without using any ordinal \(\gamma\) that is not in the set......."Boboris02 (talk) 19:10, March 12, 2017 (UTC) Would someone please explain to me what's the point of this discussion, given that LittlePeng already explained in Boboris' blog why the sigma notation is ill-defined (and as far as I can tell - none of LP's objections there were resolved)? It seems to me that it makes little sense to debate the finer points of a notation or ask "how strong is it" when the notation itself doesn't have a working definition. Sorry if this sounds harsh, but I'm asking an honest question here: How does discussing something which is ill-defined, get us anywhere? This question is directed mostly at LittlePeng (since he is the only one who said directly that the notation is ill-defined) PsiCubed2 (talk) 06:46, April 2, 2017 (UTC) :Firstly, Boris has never really acknowledged that this notation is ill-defined. In his eyes, the only problem is that his descriptions do not match exactly what he means. :That given, the reason (or, one of the reasons) the analysis page was created was so that we (mostly me and Deedlit) could take a look at what the notation was intended to do and help Boris define it formally. Unfortunately, neither the list of equalities nor Boris's attempts at explaining some of them have made me understand in the slightest how this notation really works. :As for my objections being resovled: you are mostly right about that - the only thing which was resolved is the first part of the definition (the one before \(L(\alpha)\) is defined), but I still claim the definition of \(L(\alpha)\) is wrong and Boris still doesn't seem to accept it no matter how many times I try to convince him. :So in short: it was analysed and discussed in order to fix the definition, but that ended up nowhere. LittlePeng9 (talk) 08:50, April 2, 2017 (UTC)